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fa-spin"></i></span></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h1><blockquote>
<p>题目链接：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/">236. 二叉树的最近公共祖先</a></p>
</blockquote>
<p>给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。</p>
<p>百度百科中最近公共祖先的定义为：“对于有根树 T 的两个节点 p、q，最近公共祖先表示为一个节点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”</p>
<p><strong>示例1：</strong></p>
<p><img src="https://geekwolfman-blog.oss-cn-chengdu.aliyuncs.com/articles/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88/01.png" alt="img"></p>
<figure class="highlight sql"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：root <span class="operator">=</span> [<span class="number">3</span>,<span class="number">5</span>,<span class="number">1</span>,<span class="number">6</span>,<span class="number">2</span>,<span class="number">0</span>,<span class="number">8</span>,<span class="keyword">null</span>,<span class="keyword">null</span>,<span class="number">7</span>,<span class="number">4</span>], p <span class="operator">=</span> <span class="number">5</span>, q <span class="operator">=</span> <span class="number">1</span></span><br><span class="line">输出：<span class="number">3</span></span><br><span class="line">解释：节点 <span class="number">5</span> 和节点 <span class="number">1</span> 的最近公共祖先是节点 <span class="number">3</span> 。</span><br></pre></td></tr></table></figure>

<p><strong>示例2：</strong></p>
<p><img src="https://geekwolfman-blog.oss-cn-chengdu.aliyuncs.com/articles/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88/02.png" alt="img"></p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4</span><br><span class="line">输出：5</span><br><span class="line">解释：节点 5 和节点 4 的最近公共祖先是节点 5 。因为根据定义最近公共祖先节点可以为节点本身。</span><br></pre></td></tr></table></figure>

<p><strong>示例3：</strong></p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：root = [1,2], p = 1, q = 2</span><br><span class="line">输出：1</span><br></pre></td></tr></table></figure>

<p>提示：</p>
<ul>
<li>树中节点数目在范围 [2, 105] 内。</li>
<li>-109 &lt;&#x3D; Node.val &lt;&#x3D; 109</li>
<li>所有 Node.val 互不相同 。</li>
<li>p !&#x3D; q</li>
<li>p 和 q 均存在于给定的二叉树中。</li>
</ul>
<h1 id="我的题解"><a href="#我的题解" class="headerlink" title="我的题解"></a>我的题解</h1><h2 id="方法一：层序遍历"><a href="#方法一：层序遍历" class="headerlink" title="方法一：层序遍历"></a><strong>方法一：层序遍历</strong></h2><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a><strong>思路</strong></h3><p>一个非常直观的思路是，从p、q两个节点向上找，找到第一个公共节点，即是最近公共祖先节点。因此我们先层序遍历，记录当前节点的前一节点，再向上查找第一个公共节点，代码如下：</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a><strong>代码</strong></h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    <span class="keyword">public</span> TreeNode <span class="title function_">lowestCommonAncestor</span><span class="params">(TreeNode root, TreeNode p, TreeNode q)</span> &#123;</span><br><span class="line">        Queue&lt;TreeNode&gt; queue = <span class="keyword">new</span> <span class="title class_">ArrayDeque</span>&lt;&gt;();</span><br><span class="line">        queue.offer(root);</span><br><span class="line">        <span class="type">TreeNode</span> <span class="variable">tp</span> <span class="operator">=</span> <span class="literal">null</span>, tq = <span class="literal">null</span>;</span><br><span class="line">        HashMap&lt;TreeNode, TreeNode&gt; map = <span class="keyword">new</span> <span class="title class_">HashMap</span>&lt;&gt;();</span><br><span class="line">        <span class="keyword">while</span> (!queue.isEmpty() &amp;&amp; (tp == <span class="literal">null</span> || tq == <span class="literal">null</span>)) &#123;</span><br><span class="line">            <span class="type">TreeNode</span> <span class="variable">poll</span> <span class="operator">=</span> queue.poll();</span><br><span class="line">            <span class="keyword">if</span> (poll.val == p.val) &#123;</span><br><span class="line">                tp = poll;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">if</span> (poll.val == q.val) &#123;</span><br><span class="line">                tq = poll;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">if</span> (poll.left != <span class="literal">null</span>) &#123;</span><br><span class="line">                queue.offer(poll.left);</span><br><span class="line">                map.put(poll.left, poll);</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">if</span> (poll.right != <span class="literal">null</span>) &#123;</span><br><span class="line">                queue.offer(poll.right);</span><br><span class="line">                map.put(poll.right, poll);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        HashSet&lt;TreeNode&gt; set = <span class="keyword">new</span> <span class="title class_">HashSet</span>&lt;&gt;();</span><br><span class="line">        <span class="keyword">while</span> (tp != <span class="literal">null</span>) &#123;</span><br><span class="line">            set.add(tp);</span><br><span class="line">            tp = map.get(tp);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">while</span> (tq != <span class="literal">null</span> &amp;&amp; !set.contains(tq)) &#123;</span><br><span class="line">            tq = map.get(tq);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> tq;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h3 id="结果"><a href="#结果" class="headerlink" title="结果"></a><strong>结果</strong></h3><p>执行用时：12 ms, 在所有 Java 提交中击败了6.96%的用户</p>
<p>内存消耗：42 MB, 在所有 Java 提交中击败了98.21%的用户</p>
<h2 id="方法二：深度遍历"><a href="#方法二：深度遍历" class="headerlink" title="方法二：深度遍历"></a><strong>方法二：深度遍历</strong></h2><h3 id="思路-1"><a href="#思路-1" class="headerlink" title="思路"></a><strong>思路</strong></h3><p>一般对于这种题目，我们首先考虑最基本的情况：</p>
<ul>
<li>p、q位于root左右两边，那当前节点就是最近公共祖先</li>
<li>root&#x3D;&#x3D;p，q可能位于root左边或右边，那么当前节点就是最近公共祖先</li>
<li>root&#x3D;&#x3D;q，p可能位于root左边或右边，那么当前节点就是最近公共祖先</li>
</ul>
<p>因此，采用后序遍历：</p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a><strong>代码</strong></h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    <span class="keyword">public</span> TreeNode <span class="title function_">lowestCommonAncestor</span><span class="params">(TreeNode root, TreeNode p, TreeNode q)</span> &#123;</span><br><span class="line">        <span class="keyword">if</span> (root == <span class="literal">null</span> || root.val == p.val || root.val == q.val) &#123;</span><br><span class="line">            <span class="keyword">return</span> root;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="type">TreeNode</span> <span class="variable">left</span> <span class="operator">=</span> lowestCommonAncestor(root.left, p, q);</span><br><span class="line">        <span class="type">TreeNode</span> <span class="variable">right</span> <span class="operator">=</span> lowestCommonAncestor(root.right, p, q);</span><br><span class="line">        <span class="keyword">if</span> (left != <span class="literal">null</span> &amp;&amp; right != <span class="literal">null</span>) &#123;</span><br><span class="line">            <span class="keyword">return</span> root;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> left != <span class="literal">null</span> ? left : right;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h3 id="结果-1"><a href="#结果-1" class="headerlink" title="结果"></a><strong>结果</strong></h3><p>执行用时：6 ms, 在所有 Java 提交中击败了99.99%的用户</p>
<p>内存消耗：42.9 MB, 在所有 Java 提交中击败了39.74%的用户</p>
</article><div class="post-copyright"><div class="post-copyright__author"><span class="post-copyright-meta">文章作者: </span><span class="post-copyright-info"><a href>狼族少年、血狼</a></span></div><div class="post-copyright__type"><span class="post-copyright-meta">文章链接: </span><span class="post-copyright-info"><a href="https://geekwolfman.github.io/2023/04/18/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88.html">https://geekwolfman.github.io/2023/04/18/二叉树的最近公共祖先.html</a></span></div><div class="post-copyright__notice"><span class="post-copyright-meta">版权声明: </span><span class="post-copyright-info">本博客所有文章除特别声明外，均采用 <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" target="_blank">CC BY-NC-SA 4.0</a> 许可协议。转载请注明来自 <a href="https://geekwolfman.github.io" target="_blank">狼族少年、血狼</a>！</span></div></div><div class="tag_share"><div class="post-meta__tag-list"><a class="post-meta__tags" href="/tags/%E7%AE%97%E6%B3%95%E9%A2%98/">算法题</a></div><div 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class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#%E9%A2%98%E7%9B%AE%E6%8F%8F%E8%BF%B0"><span class="toc-number">1.</span> <span class="toc-text">题目描述</span></a></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E6%88%91%E7%9A%84%E9%A2%98%E8%A7%A3"><span class="toc-number">2.</span> <span class="toc-text">我的题解</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%96%B9%E6%B3%95%E4%B8%80%EF%BC%9A%E5%B1%82%E5%BA%8F%E9%81%8D%E5%8E%86"><span class="toc-number">2.1.</span> <span class="toc-text">方法一：层序遍历</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF"><span class="toc-number">2.1.1.</span> <span class="toc-text">思路</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E4%BB%A3%E7%A0%81"><span class="toc-number">2.1.2.</span> <span class="toc-text">代码</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E7%BB%93%E6%9E%9C"><span class="toc-number">2.1.3.</span> <span class="toc-text">结果</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%96%B9%E6%B3%95%E4%BA%8C%EF%BC%9A%E6%B7%B1%E5%BA%A6%E9%81%8D%E5%8E%86"><span class="toc-number">2.2.</span> <span class="toc-text">方法二：深度遍历</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-1"><span class="toc-number">2.2.1.</span> <span class="toc-text">思路</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E4%BB%A3%E7%A0%81-1"><span class="toc-number">2.2.2.</span> <span class="toc-text">代码</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E7%BB%93%E6%9E%9C-1"><span class="toc-number">2.2.3.</span> <span class="toc-text">结果</span></a></li></ol></li></ol></li></ol></div></div></div></div></main><footer id="footer"><div id="footer-wrap"><div class="copyright">&copy;2020 - 2023 By 狼族少年、血狼</div><div 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